1. The major product of the following reaction is
 Ans
PhS- is a strong Na- ( also, note the presence of aprotic solvent => SN2 will occure leading to inversion(Walden) - F can not be substituted by PhS- at –NO2 is present at the “m” position
 2. Hyperconjugation involves overlap of the following orbitals
 Ans
By the definition of hyper conjugation, it involves the overlap of s & P orbital
3. Aqueous solution of Na2S2O3 on reaction with Cl2 gives
(a) Na2S4O6 (b) NaHSO4 (c ) NaCl (d) NaOH
Ans
 4. Native silver metal forms a water complex with a dilute aqueous solution of NaCN in the presence of
(A) nitrogen (B) oxygen (C) carbon dioxide (D) argon
Ans
 5. 2.5 mL of 2/5 M weak monoacidic base (Kb=1X10-12 at 25o C is titrated with 2/15 M HCI in water at 25 oC. The concentration of H+ at equivalence point is
 Ans
Calculate the vol. of HCl to reach the equivalence point
 => net vol of solution at the equation point = Vbase + VHCl =2.5 + 7.5 = 10 ml
 please note the above solution is incorrect(why?)
 6. Under the same reaction condition, initial concentration of 1.386 mol dm-3 of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio (A) 0.5 mol-1 dm3 (B) 1.0 mol dm-3 (c) 1.5 mol dm-3 (d) 2.0 mol-1 dm3
Ans
first order: half life:
 
7. The correct statement(s) concerning the structures E, F and G is (are)
 (A) E, F and G are resonance structures (B) E, F and K, G are tautomers (C) F and G are geometrical isomers (D) F and G are diastereomers
Ans
F is clearly the tautomer (enol form) of E G is Geometric isomer(trans form) of F => E & G are also tautomers ( geometric isomers are also diastereomers) 8. The correct statement(s) about the compound given below is (are)
 (A) The compound is optically active (B) The compound possesses centre of symmetry (C) The compound possesses plane of symmetry (D) The compound possesses axis of symmetry Ans
 center of symmetry: for every atom with coordinates x, y, z there exists a similar atom with coordinate -x, -y, -z with inversion center being the origin of the coordinates. Plane of symmetry: plane which divides the molecule into halves which are mirror image of each other. Axis of symmetry: If a molecule is rotated about an imaginary axis by an angle of 360o/n & arrives at an arrangement indistinguishable from the original, the axis is n - fold axis of symmetry.
9. A gas described by van der Waals equation
(A) Behaves similar to an ideal gas in the limit of large molar volumes
(B) Behaves similar to an ideal gas in the limit of large pressures (C) Is characterized by van der walls coefficients that are dependent on the identity of the gas but are independent of the temperature (D) Has the pressure that is lower than the pressure exerted by the same gas behaving ideally Ans.  As a, b are the characteristic constants of a real gas(independent of T) since there are attractive forces between gas molecules, real gas pressure is always less than ideal gas pressure.
10. A solution of colourless salt.H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H-is (are)
(A) NH4NO3 (B) NH4NO2
(C ) NH4CI (D) (NH4)2SO4
Ans. Note: all of the given components: NH4OH; NH4NO2; NH4Cl; (NH4)2SO4 will react with NaOH to give NH3 gas. 
11. STATEMENT-1: The plot of atomic number (y-axis) versus number of neutrons (x-axis) for stable nuelei shows a curvature towards x-axis from the line of 450 slope as the atomic number is increased. And STATEMENT-2: Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is False Ans.
 In heavy nuclei’s n/p ratio > 1As Z increases, attractive nuclear tends to saturate ( saturation property of nuclear force) and after this, proton-proton repulsion tends to dominate leading to
12. STATEMENT-1: For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero
And
STATEMENT-2: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is False
Ans. for a chemical reaction at equilibrium
 but reactants & products to achieve minimum gibbs energy
 13. STATEMENT-1: Bromobenzene upon reaction with Br2/Fe gives 1, 4-dibromobenzene as the major product.
And
STATEMENT-2: In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile.
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is False
Ans. -Br is weakly -O, p directing group due to +me effort. Inductive effect (-I) of halogens is stronger then their +M effect. This makes halogens as weakly deactivating group but incoming electrophile is directed to ‘O’ & ‘P’ position because of +Me effect though this effect is weakand.
14. STATEMENT-1: Pb2+ compounds are stronger oxidizing agents than Sn4+compounds.
And
STATEMENT-2: The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect'
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1 (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is False
Ans. Sn4+is max stable than SN2 [SN2+ ions in the solution are good reducing agents]
Pb2+ is more stable than Pb4+ => Pb4+ compounds are stranger oxidizing agents than Sn4+ (true) As we move down the “C” group, higher oxidation state (+4) becomes more stable due to “inert-pair” effect but Pb is an exceptional case. (true)
Paragraph for Question Nos. 15 to 17 Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day to day life. One of its examples is the use of ethylene glycol and water mixture as anti- freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given: Freezing point depression constant of water Freezing point depression constant of ethanol Boiling point elevation constant of water Boiling point elevation constant of ethanol Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol-1 Molecular weight of ethanol = 46 g mol-1 In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non- dissociative. 15. The freezing point of the solution M is
(A) 268.7 K (B) 268.5 K (C) 150.9 K (D) 268.7 K
Ans. Note; XEthanol = 0.9(solvent); Xwater = 0.1 (solute)
=> freezing point of solution = 155.7 - 4.8 = 150.9 k
16. The vapour pressure of the solution M is
(A) 39.3 mm Hg (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mm Hg
Ans.
To get vapour pressure of the solution M: [note: wter is assumed to be non – volatile]
17. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling point of this solution is
(A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K
Ans. XEthanol = 0.9(solute); Xwater = 0.1 (solvent)
Paragraph for Question Nos. 18 to 20
In the following reaction sequence, products I, J and L are formed. K represents a reagent.
 Ans.
Paragraph for Question Nos. 21 to 23
There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridization easily explains the ease of sigma donation capability if NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorous.
21. Among the following, the correct statement is
(A) Phosphates have no biological significance in humans (B) Between nitrates and phosphates, phosphates are less abundant in earth's crust (C) Between nitrates and phosphates, nitrates are less abundant in earth's crust (D) Oxidation of nitrates is possible in soil
Ans. Microbes will reduce nitrates in earth’s crust. So, nitrates are less abundant in earth’s crust.
22. Among the following, the correct statement is
(A) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional. (B) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electrons occupies sp3 orbital and is more directional. (C) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electrons occupies sp3 orbital and is more directional. (D) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional. Ans. NH3 is stronger electron pair donor because N -atom is having the smaller size & electron density of the ion pair of e-s present in one of the SP3 hybrids on N - atom is more concentrated over a smaller region.
23. White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a
(A) dimerization reaction (B) disproportionation reaction (C) Condensation reaction (D) Precipitation reaction
Ans.  (disproportionation reaction) |
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Solutions: Chemistry Paper 1 IITJEE 2008 |
deviation from ideal gas
= 1.86 K kg mol-1
